University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 3 - Motion in Two or Three Dimensions - Problems - Exercises - Page 98: 3.70


$h = \frac{2gv^2}{a^2}$

Work Step by Step

Since $v$ and $a$ are in opposite directions, we can write an expression for the horizontal position $x$ as follows: $x = v~t - \frac{1}{2}at^2$ Since the firecracker lands directly below the student, we can let $x = 0$. $x = v~t - \frac{1}{2}at^2 = 0$ $t = \frac{2v}{a}$ We can find an expression for the height $h$. $h = \frac{1}{2}gt^2$ $h = (\frac{g}{2})(\frac{2v}{a})^2$ $h = (\frac{g}{2})(\frac{4v^2}{a^2})$ $h = \frac{2gv^2}{a^2}$
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